STRUCTURE AND BINDING OF H3 AND He3
By Prof. L. Kaliambos (Natural Philosopher in New Energy) 25 February 2019 It is indeed unfortunate that the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established laws of electromagnetism in favour of wrong nuclear theories like the wrong meson theory (1935) and the invalid quantum chromodynamics (1973) which could not lead to the correct nuclear structure . Under this physics crisis I was based on the charged quarks discovered by Gell-Mann and Zweig for publication of my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003) which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons Note that the paper of my discovery of nuclear force and structure presented also at a nuclear conference held at NCSR "Demokritos" (2002), where I showed that the 9 extra charged quarks in proton and the 12 ones in neutron are responsible for the charge distributions in nucleons which give the nuclear structure. Today it is well-known that the structures and binding energies of nuclei are based not on invalid nuclear theories but on the well-established laws of electromagnetism acting at a distance. (Quantum Entanglement rejects Einstein). Then for understanding the simplest nuclear structure based on strong electromagnetic forces one can see my DEUTERON STRUCTURE AND BINDING . In other words the simplest explanation of the deuteron structure and binding is able to tell us how the charge distributions of two spinning nucleons interact electromagnetically with parallel spin ( S = 1) for giving the nuclear binding and force. Also for understanding the bonds of stable and unstable nuclei one can see my paper SRUCTURE AND BINDING OF H4 AND He6. In the following diagram of He4 one sees that the spinning nucleons in He4 form a simple rectangle with S=0 which explains the stability of alpha particles, because the neutron n1 forms two bonds with p1 and p2 and also the neutron n2 forms two bonds with p1 and p2 . Stable Helium -4 n2(-1/2)..p2 (- 1/2) p1(+1/2)..n1(+1/2) However using the diagram of the He-4 you can draw the diagram of H-3 (Tritium), which is an unstable nucleus , because the two neutrons n1 and n2 form single bonds with the proton p1. Diagram of Unstable H-3 (Tritium) n2(-1/2).............. p1(+1/2)..n1(+1/2) Under this condition Tritium is a radioactive isotope of hydrogen. The nucleus of tritium (sometimes called a triton) ..... Tritium is an isotope of hydrogen, which allows it to readily bind to hydroxyl radicals, forming tritiated water (HTO), and to carbon atoms. Binding energy‎: ‎8,481.821± 0.004 keV Decay mode‎: ‎Decay energy‎ (‎MeV‎) Decay products‎: ‎3He Isotope mass‎: ‎3.0160492 u On the other hand in the stable Helium -3 the neutron n1 forms two bonds with p1 add p2 as Diagram of Stable Helium- 3 ...............p2 (- 1/2) p1(+1/2)..n1(+1/2) Under this condition Helium-3 (3He )is a light, non-radioactive isotope of helium with two protons and one neutron (common helium having two protons and two neutrons). ... Other than protium (ordinary hydrogen), helium-3 is the only stable isotope of any element with more protons than neutrons. Isotope mass‎: ‎3.0160293 u Neutrons‎: ‎1 Parent isotopes‎: ‎3H (beta decay of tritium) Natural abundance‎: ‎0.000137% (% He on Eart... Category:Fundamental physics concepts